Upper bound and lower bound

  • Thread starter l46kok
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  • #1
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Homework Statement



1.
f(n) = n - 100
g(n) = n - 200

2.
f(n) = log(2n)
g(n) = log(3n)

n >= 0 in all cases
Find out if f(n) is an upperbound, lowerbound or both of g(n)

Homework Equations





The Attempt at a Solution



in case of 1, f(n) has to be an upperbound of g(n) because when graphed together, f(n) has to be an upperbound of g(n).

For 2, solution does not exist at n = 0. Otherwise, f(n) is a lower bound of g(n). Does this mean that f(n) is a lower bound or both?
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement



1.
f(n) = n - 100
g(n) = n - 200

2.
f(n) = log(2n)
g(n) = log(3n)

n >= 0 in all cases
Find out if f(n) is an upperbound, lowerbound or both of g(n)

Homework Equations





The Attempt at a Solution



in case of 1, f(n) has to be an upperbound of g(n) because when graphed together, f(n) has to be an upperbound of g(n).
Too vague. What you should say is "200> 100 so -100> -200 and n- 100> n- 200 for all n. Since f(n)> g(n) for all n, f is an upper bound of g."

For 2, solution does not exist at n = 0. Otherwise, f(n) is a lower bound of g(n). Does this mean that f(n) is a lower bound or both?
Is suspect that should not be "[itex]n\ge 0[/itex] for both cases" but only n> 0 for the second. As you point out, ln(0) is not defined so the problem makes no sense for n= 0.

ln(2)< ln(3) so ln(n)+ ln(2)< ln(n)+ ln(3) for all n> 0. ln(2n)< ln(3n) for all n> 0.
 

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